a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

giải phần còn lại giúp mình được ko?

Bài 3:

a:

ĐKXĐ: x>=-5

\(x^2-7x=6\sqrt{x+5}-30\)

=>\(x^2-4x-3x+12=6\sqrt{x+5}-18\)

=>\(\left(x-4\right)\left(x-3\right)=6\left(\sqrt{x+5}-3\right)\)

=>\(\left(x-4\right)\left(x-3\right)=6\cdot\frac{x+5-9}{\sqrt{x+5}+3}\)

=>\(\left(x-4\right)\left(x-3-\frac{6}{\sqrt{x+5}+3}\right)=0\)

=>x-4=0

=>x=4(nhận)

Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{x+4\sqrt{x}+4}=5x+2\)

=>\(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

=>\(5x+2=\sqrt{x}+2\)

=>\(5x-\sqrt{x}=0\)

=>\(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}=0\\ 5\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ \sqrt{x}=\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\frac{1}{25}\left(nhận\right)\end{array}\right.\)

b: \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)

=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\) =4

=>|x-1|+|x+2|=4(1)

TH1: x<-2

(1) sẽ trở thành: -x-2+1-x=4

=>-2x-1=4

=>-2x=5

=>x=-5/2(nhận)

TH2: -2<=x<1

(1) sẽ trở thành: x+2+1-x=4

=>3=4(vô lý)

TH3: x>=1

(1) sẽ trở thành: x+2+x-1=4

=>2x+1=4

=>2x=3

=>x=3/2(nhận)

c: ĐKXĐ: x>=1

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)

=>\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)

=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\)

=>\(\left|\sqrt{x-1}-1\right|=2-\sqrt{x-1}-1=1-\sqrt{x-1}\)

=>\(\sqrt{x-1}-1\le0\)

=>\(\sqrt{x-1}\le1\)

=>0<=x-1<=1

=>1<=x<=2

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

x^2(-2x^2+x-2)=0

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

Câu 1:

\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)

\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)

\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)

\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)

\(\Leftrightarrow50x-16=0\)

\(\Leftrightarrow50x=16\)

\(\Leftrightarrow x=\dfrac{8}{25}\)

Câu 2 :

\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)

<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)

<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)

<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x

<=> 11x+27 = 26x -5

<=> ( 26x - 5 ) - ( 11x + 27 ) = 0

<=> 15x - 32 = 0

<=> 15x = 32

<=> x = \(\dfrac{32}{15}\)

a: \(\left(2x^2+5x-2\right)\left(2x^2-4x+3\right)\)

\(=4x^4-8x^3+6x^2+10x^3-20x^2+15x-4x^2+8x-6\)

\(=4x^4+2x^3-18x^2+23x-6\)

b: \(\left(2x-3\right)\left(3x-2\right)-3x\left(2x-5\right)\)

\(=6x^2-4x-9x+6-6x^2+15x\)

=2x+6

c: \(\left(x-1\right)\left(x^2+x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)

\(=x^3-1-\left(x^3+1\right)\)

\(=x^3-1-x^3-1=-2\)

d: \(\left(x^2+x-1\right)\cdot\left(x^2-x+1\right)=\left(x^2\right)^2-\left(x-1\right)^2\)

\(=x^4-\left(x^2-2x+1\right)=x^4-x^2+2x-1\)

e: Sửa đề: \(\left(2x+3y\right)^2-\left(2x-3y\right)^2-12xy\)

=(2x+3y+2x-3y)(2x+3y-2x+3y)-12xy

=4x*6y-12xy

=24xy-12xy

=12xy

f: \(\left(x^2-4x\right)\left(5+2x-x^2\right)\)

\(=5x^2+2x^3-x^4-20x-8x^2+4x^3=-x^4+6x^3-3x^2-20x\)

1) 2x – (3 – 5x) = 4( x +3)

<=>2x-3+5x=4x+12

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2

j: (2x-1)(3x+1)-(4-3x)(3-2x)=3

=>(2x-1)(3x+1)-(3x-4)(2x-3)=3

=>\(6x^2+2x-3x-1-\left(6x^2-9x-8x+12\right)=3\)

=>\(6x^2-x-1-\left(6x^2-17x+12\right)=3\)

=>\(6x^2-x-1-6x^2+17x-12=3\)

=>16x-13=3

=>16x=16

=>x=1

k: \(\left(2x+1\right)\left(x+3\right)-\left(x-5\right)\left(2x+7\right)=8\)

=>\(2x^2+6x+x+3-\left(2x^2+7x-10x-35\right)=8\)

=>\(2x^2+7x+3-2x^2-7x+10x+35=8\)

=>10x+38=8

=>10x=-30

=>x=-3

m: 2(3x-1)(2x+5)-6(2x-1)(x+2)=-6

=>\(2\left(6x^2+15x-2x-5\right)-6\left(2x^2+4x-x-2\right)=-6\)

=>\(2\left(6x^2+13x-5\right)-6\left(2x^2+3x-2\right)=-6\)

=>\(12x^2+26x-10-12x^2-18x+12=-6\)

=>8x+2=-6

=>8x=-8

=>x=-1

A) 3x² - x(3x - 5) = 9

3x² - 3x² + 5x = 9

5x = 9

x = 9/5

--------------------

B) 5x² + 9x - 2 = 0

5x² + 10x - x - 2 = 0

(5x² + 10x) - (x + 2) = 0

5x(x + 2) - (x + 2) = 0

(x + 2)(5x - 1) = 0

x + 2 = 0 hoặc 5x - 1 = 0

*) x + 2 = 0

x = -2

*) 5x - 1 = 0

5x = 1

x = 1/5

Vậy x = -2; x = 1/5

---------------------

D) 4(5 - 3x) = 5x - 5

20 - 12x = 5x - 5

-12x - 5x = -5 - 20

-17x = -25

x = 25/17

--------------------

E) 2x² - 11x + 14 = 0

2x² - 4x - 7x + 14 = 0

(2x² - 4x) - (7x - 14) = 0

2x(x - 2) - 7(x - 2) = 0

(x - 2)(2x - 7) = 0

x - 2 = 0 hoặc 2x - 7 = 0

*) x - 2 = 0

x = 2

*) 2x - 7 = 0

2x = 7

x = 7/2

Vậy x = 2; x = 7/2

Câu C và F ghi đề bằng công thức đúng lại em